Weeks 1-3 ReviewFall 2006 Mth 256Bent E. PetersenThis review deals with recipes for solving first order ordinary differential equations as studied inthe first three weeks of Mth 256, Fall 2006. The theoretical parts of the course are not reviewedhere. The problems listed here do not represent every type of problem we discussed.Answers indicated below were found mostly by Maple (for fun) and so may differ in some inessen-tial way from the answers that you find. While I copy and paste the LATEX version of the answersas provided by Maple, I can’t help but clean up Maple’s LATEX expressions a bit. Undoubtably Iintroduced a few errors, so you should rely more on yourself than on the provided “answers.”Enjoy!1 Picard IterationProblem 1. The initial value problemdydt= t2+ y2, y(0) = 1can be solved in terms of certain Bessel functions. If we expand the solution in a Taylor series weobtainy(t) = 1+ t + t2+43t3+76t4+65t5+3730t6+404315t7+369280t8+ ···If you perform three Picard iterations (starting with y0(t) = 1) what do you obtain? How does yourresult compare with the solution given above?Answer:1+t +t2+43t3+56t4+815t5+2990t6+47315t7+ ···2 Linear First Order ODEProblem 2. Solve the initial value problemdydt+ tan(t) y = tan(t), y(0) = 3.Mth 256 Weeks 1-3 ReviewAnswer:y(t) = 1+ 2 cos(t)Problem 3. Solve the initial value problemdydt+yt= sin(t), y(π) = 2.Answer:y(t) =sin(t) −cos(t) t +πtProblem 4.dydt=yt −3+t2Answer:y(t) = (t −3)t22+ 3t + 9 log(t −3) +CProblem 5.dydt= cot(t)y+ sin(t)Answer:y(t) = (t +C)sin(t)Problem 6.dydt=yt+ sin(log(t))Answer:y(t) = −t cos(log(t)) +CtProblem 7. A certain radioactive substance decays to 85 % of its original mass in 36 hours.Find the half–life.Answer:153.54 hoursBent E. Petersen Page 2 Fall 2006Mth 256 Weeks 1-3 Review3 Newton’s Law of CoolingProblem 8. A cup of hot coffee initially at temperature 134◦F is brought into a room oftemperature A (the ambient). The coffee begins to cool down and, of course the room warms up abit. However, the heat capacity of the room is so large compared to the cup of coffee, that we mayassume A remains constant. After 2 minutes the coffee is observed to have the temperature 100◦K.Another minute later the coffee is observed to have the temperature 90◦F. Deduce the temperatureof the room.Answer:A = 65.64◦F.Problem 9. A cup of coffee initially at 190◦F is brought into a room at 65◦F. After 2 minutesthe temperature of the coffee is 145◦F. Predict the temperature of the coffee an additional minutelater.Answer:129◦FProblem 10. Consider an insulated box with internal temperature T. Assume that the ambi-ent (external) temperature A is changing linearly (for a while at least), sayA = A0+ A1twhere A0and A1are constants, and t is time. According to Netwon’s law of cooling we havedTdt= −k (T −A)where k is a constant depending on the insulation of the box. Find the temperature T(t) in termsof t, A0, A1and k. (Do not neglect the arbitrary constant.)Answer:T(t) = A0+ A1t −1k+Ce−ktBent E. Petersen Page 3 Fall 2006Mth 256 Weeks 1-3 Review4 Mixing ProblemsProblem 11. A brine solution consisting of 0.06 oz/gal salt dissolved in water flows into alarge tank at the rate 3.0 gal/min. The solution inside the tank is kept well-mixed and flows out ofthe tank at the rate 2.0 gal/min. If the tank initially contains 50.0 gal of brine of concentration 0.03oz/gal determine the amount of salt in the tank after t minutes. When will the concentration of saltin the tank reach 0.05 oz/gal? Assume the tank is so large that it does not overflow.Answer:22.1 minProblem 12. A 200 L tank initially contains 100 L of brine of concentration 3 g/L salt (i.e.,3 grams salt per liter water). Brine of concentration 5 g/L salt runs into the tank at 8 L/min. Thewell–mixed solution is drawn off at the rate 6 L/min. Find the concentration of salt in the solutionin the tank at the moment that the tank begins to overflow.Answer:4.875 g/LProblem 13. A 100 gal tank initially contains 20 gal of brine of concentration 0.24 oz/galsalt. Brine of concentration 0.18 oz/gal flows into the tank at 3 gal/min and the well-mixed solutionis drawn off at the rate of 1 gal/min. Find the amount of salt in the tank at the very moment that itbegins to overflow.Answer:18.537 oz5 Separable First Order ODEProblem 14. For a body of mass m (say a person) falling near the surface of the earth wemay assume the acceleration of gravity g is a constant. Also in many cases we may assume thedrag is proportional to the square of the velocity. Thus the equation of motion ismdvdt= mg−kv2.Bent E. Petersen Page 4 Fall 2006Mth 256 Weeks 1-3 ReviewHere v is the downward velocity. If we reparametrize this equation by the distance y fallen (try it)we obtainmvdvdy= mg−kv2.Assuming the body falls from rest find the velocity in terms of the distance fallen. From yoursolution determine the terminal, ultimate or limiting velocity (symbolically) if you fall a verylong way. How far does a 60 kg person have to fall to reach 95 % of the terminal velocity ifk = 0.60 kg/m? (Note k depends on the person’s aspect. For feet or head first we may have roughlyk = 0.06.)Note you do not need the acceleration of gravity g to answer the questions above, but you do tocompute the actual speed. Use g = 9.80 m/sec2to compute the 95 % of terminal velocity achievedby our intrepid aeronaut.Answers:v =mgk1/21−e−2ky/m1/2,mgk1/2,and 116.4 m with a speed of 29.7 m/sec, or about 66.5 mph.Problem 15. Solve the initial value problemdydx=xy+ yx, y(1) = −2.Answer:y = −xex−1Problem 16. Solve the initial value problemdpdt= e2tp(1− p), p(0) = 0.5.Findlimt→∞p(t).Answer:p(t) =11+ e(1−e2t)/2, The limit is 1.Bent E. Petersen Page 5 Fall 2006Mth 256 Weeks 1-3 Review6 Bernoulli ODEProblem 17. Solve the ordinary differential equationy3dydx+y4x=tan(x)x4.Answerx4y4+ 4 log(cos(x)) = CProblem 18. Solve the ordinary differential equationxdydx+ 2y+√y log(x) = 0.Answer√y+12log(x) −12=Cx7 Homogeneous First Order ODE and related substitutionsProblem 19. Use the substitution y = x2w to solve the ordinary differential equation (nothomogeneous)dydx=2y2+ x3xy.Answer:y2+ 2x3+Cx4= 0Problem 20. Solve the initial value problemdydx=4x2+ y22xy, y(2) = 2.Answer:y(x) =4x2−6x1/2Bent E. Petersen Page 6 Fall 2006Mth 256 Weeks 1-3 ReviewProblem 21. Solve the initial value problemdydx= y(1+ x2) −(1+ x2), y(0) = 2.Answer:y(x) = 1+ ex+x33.(You can also solve this